Close. Displacement currents do not create heat, only conduction currents.
Matt Watts
Re: The Energy Machine of Joseph Newman 2017
« Reply #1525, on March 21st, 2018, 08:44 PM »
Close. Displacement currents do not create heat, only conduction currents.
So essentially a low-resistance battery would 'outrun' the rotational rate of the rotor, allowing the battery to continue pumping voltage into the coil and accelerating the rotor, whereas a high-resistance battery has a voltage dip which allows the induced momentum of the rotor (after the first 15 - 20 degrees of rotation) to induce current in the same direction as the applied current.
Resistance is the key to this device, or should I say impedance.
Is kicking our ass. We need to flip things around so that as resistance goes down, voltage AND current goes up. We need conductance, not resistance.
In fact, it's a mathematical certainty... remember, all interactions must lose to heat at least 1/c2 of the energy transferred.
so there for it all depends on the system. if your going from dielectric from magnetic back to dielectric you can do a much better job at transferring the energy.
Magic sQuar 2O18
...
believe...whatever you want to believe.....you stay in wonderland
Magic sQuar 2O18
...
believe...whatever you want to believe.....you stay in wonderland
Work with me here Cycle. We need to understand something very much overlooked about a resistor and resistance in general.
When Russ has too low of resistance with his batteries, the batteries themselves lock the rotor from turning at the rate it is capable of. This is a resistance inside of the batteries, in respect of the coil resistance. Now if we could make that internal battery resistance go sky-high for a short moment in time, the rotor would fly right out of the coil. What are we missing?
How are you checking resistance on a battery. On a typical multi meter it has its own battery that it uses for a reference, so depending on how much charge the batteries have that you are checking it would seem the resistance readings would not be accurate or vary quite a bit..
Id say this is not true
That's not to say you can't more than make up for that energy loss in creative ways (such as using a diode, a switch, an inductor and a ground connection to create an upstream 'vacuum' to pull more current than that which was transferred).
Your saying that ALL energy transfers lose 1/c2 of the energy released or transferred.
I'm confused.
Your saying that ALL energy transfers lose 1/c2 of the energy released or transferred.
Even in super conductors?
The word ALL implies. Every. I don't think thats a corect statement. Can you please point me to some readding on this so I can understand?
Thanks!
~Russ
for some upcoming tests. https://youtu.be/RxN1dtmxJcs
~Russ
That's crazy! :-)